Derivative of f ( x ) = sin x MCV4U: Calculus & Vectors While - - PDF document

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Derivative of f ( x ) = sin x MCV4U: Calculus & Vectors While - - PDF document

d e r i v a t i v e s o f t r i g o n o m e t r i c , e x p o n e n t i a l & l o g a r i t h m i c f u n c t i o n s d e r i v a t i v e s o f t r i g o n o m e t r i c , e x p o n e n t i a l & l o g a r i t h m i c f u n c t i o n s


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d e r i v a t i v e s o f t r i g o n o m e t r i c , e x p o n e n t i a l & l o g a r i t h m i c f u n c t i o n s

MCV4U: Calculus & Vectors

Derivatives of Sinusoidal Functions

(Sine and Cosine)

  • J. Garvin

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Derivative of f (x) = sin x

While we have dealt with derivatives of polynomial, radical and reciprocal functions, we have not yet dealt with the derivatives of sinusoidal functions like f (x) = sin x. We might calculate the rate of change every π

3 radians and

compare it to the graph of sine. From the graph, it appears that the rate of change of the sine function is the cosine function.

  • J. Garvin — Derivatives of Sinusoidal Functions

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Derivative of f (x) = sin x

To establish the derivative of the sine function, use the limit definition of the derivative. f ′(x) = lim

h→0

sin(x + h) − sin x h Use the angle sum identity for sine. f ′(x) = lim

h→0

sin x cos h + sin h cos x − sin x h Common factor sin x. f ′(x) = lim

h→0

sin x(cos h − 1) + sin h cos x h

  • J. Garvin — Derivatives of Sinusoidal Functions

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Derivative of f (x) = sin x

Use the multiplicative property of limits. f ′(x) = lim

h→0

sin x(cos h − 1) h + lim

h→0

sin h cos x h Since we are concerned with the limit of h, both sin x and cos x can be treated like constants outside of the limits. f ′(x) = sin x · lim

h→0

cos h − 1 h + cos x · lim

h→0

sin h h At this point, we need to establish values for the two limits above. Proofs of these limits generally rely on the Squeeze Theorem and some trigonometric inequalities, but some informal arguments should convince us of their values.

  • J. Garvin — Derivatives of Sinusoidal Functions

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Derivative of f (x) = sin x

Recall that in radian measure, arc length of a circle is given by a = rθ, where θ is the angle subtending the arc and r is the radius of the circle. Consider the unit circle below. When r = 1, a = θ. The height of a right triangle inside of the sector has a height of sin θ.

  • J. Garvin — Derivatives of Sinusoidal Functions

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Derivative of f (x) = sin x

Now consider what happens as θ → 0. The length of the arc, θ, and the height of the triangle, sin θ, become closer in value. If θ was infinitesimally small, then θ and sin θ would essentially have the same value. Their ratio, sin θ θ , would be 1. Therefore, lim

θ→0

sin θ θ = 1.

  • J. Garvin — Derivatives of Sinusoidal Functions

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Derivative of f (x) = sin x

A similar argument can be used for lim

θ→0

cos θ − 1 θ . Consider what happens to the horizontal component of a right triangle, cos θ, as θ → 0. If θ was infinitesimally small, then cos θ would essentially have the same length as the radius, 1. Therefore, cos θ − 1 = 1 − 1 = 0, and lim

θ→0

cos θ − 1 θ = 0.

  • J. Garvin — Derivatives of Sinusoidal Functions

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Derivative of f (x) = sin x

Now return to the previous definition of the derivative of f (x) = sin x and substitute these values. f ′(x) = sin x · lim

h→0

cos h − 1 h + cos x · lim

h→0

sin h h = sin x · 0 + cos x · 1 = cos x This confirms our graph earlier.

Derivative of the Sine Function

If f (x) = sin x, then f ′(x) = cos x.

  • J. Garvin — Derivatives of Sinusoidal Functions

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Derivative of f (x) = cos x

Use the same process for the derivative of f (x) = cos x. f ′(x) = lim

h→0

cos(x + h) − cos x h = lim

h→0

cos x cos h − sin x sin h − cos x h = lim

h→0

cos x(cos h − 1) − sin x sin h h = cos x lim

h→0

cos h − 1 h − sin x lim

h→0

sin h h = cos x · 0 − sin x · 1 = − sin x

Derivative of the Cosine Function

If f (x) = cos x, then f ′(x) = − sin x.

  • J. Garvin — Derivatives of Sinusoidal Functions

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Derivatives Involving Sinusoidal Functions

We can use the derivative rules developed earlier to find the derivatives of functions involving either sine or cosine.

Example

Determine the derivative of f (x) = 10 cos x + 4. Since 10 is a constant multiple, and 4 is a constant, f ′(x) = −10 sin x.

Example

Determine the derivative of f (x) = sin x − cos x. Using the difference rule for derivatives, f ′(x) = cos x − (− sin x) = cos x + sin x.

  • J. Garvin — Derivatives of Sinusoidal Functions

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Derivatives Involving Sinusoidal Functions

Example

Determine the derivative of y = cos(2x3 + 5x). Use the chain rule, where the inner function is u = 2x3 + 5x and the outer function is y = cos u.

dy dx = − sin(2x3 + 5x)(6x2 + 5)

Example

Determine the derivative of y = 7 sin3 x + 2 cos2 x. Use the chain rule and the fact that sinnx = (sin x)n.

dy dx = 21(sin x)2(cos x) + 4(cos x)(− sin x)

= 21 sin2 x cos x − 4 sin x cos x

  • J. Garvin — Derivatives of Sinusoidal Functions

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Derivatives Involving Sinusoidal Functions

Example

Determine the derivative of y = 2 sin x cos x. Use the product rule to differentiate.

dy dx = 2(cos x cos x − sin x sin x)

= 2(cos2 x − sin2 x) = 2 cos 2x An alternative solution uses the chain rule and the identity y = 2 sin x cos x = sin 2x.

dy dx = cos(2x)(2)

= 2 cos 2x

  • J. Garvin — Derivatives of Sinusoidal Functions

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Derivatives Involving Sinusoidal Functions

Example

Determine the slope of the tangent to f (x) = 2x cos x when x = 5π

3 .

Use the product rule to differentiate. f ′(x) = 2 cos x + 2x(− sin x) = 2(cos x − x sin x) Find f ′ 5π

3

  • for the slope of the tangent.

f ′ 5π

3

  • = 2
  • 1

2 − 5π 3

√ 3 2

  • = 1 + 5

√ 3π 3

  • J. Garvin — Derivatives of Sinusoidal Functions

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Derivatives Involving Sinusoidal Functions

Example

Find any values of x for which the tangent to y = 3 sin2 x, on the domain [0, 2π], has a slope of 3

2.

Using the chain rule, the derivative is dy

dx = 6 sin x cos x, or dy dx = 3 sin 2x after applying the double-angle formula.

Set dy

dx = 3 2 and solve for 2x.

3 sin 2x = 3

2

sin 2x = 1

2

2x = sin−1 1

2

  • 2x = π

6

  • J. Garvin — Derivatives of Sinusoidal Functions

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Derivatives Involving Sinusoidal Functions

Since sin 2x has a period of π, two cycles will be completed

  • n the interval [0, 2π]. Thus, there should be four values of x

for which this equation is true. The first two can be found by using 2x as the reference angle. 2x = π

6

π − 2x = π

6

x = π

12

x = π − π

6

2 = 5π

12

Find the other two by adding the period, π, to each value. x = π + π

12

x = π + 5π

12

= 13π

12

= 17π

12

  • J. Garvin — Derivatives of Sinusoidal Functions

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Derivatives Involving Sinusoidal Functions

A graph of y and dy

dx is below.

  • J. Garvin — Derivatives of Sinusoidal Functions

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Questions?

  • J. Garvin — Derivatives of Sinusoidal Functions

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